3.0 KiB
[1792] Maximum Average Pass Ratio
題目資訊
- 難度: Medium
- 標籤: Array, Greedy, Heap(Priority Queue)
- 題目連結: LeetCode
- 練習日期: 2025-09-01
題目描述
There is a school that has classes of students and each class will be having a final exam. You are given a 2D integer array classes, where classes[i] = [passi, totali]. You know beforehand that in the ith class, there are totali total students, but only passi number of students will pass the exam.
You are also given an integer extraStudents. There are another extraStudents brilliant students that are guaranteed to pass the exam of any class they are assigned to. You want to assign each of the extraStudents students to a class in a way that maximizes the average pass ratio across all the classes.
The pass ratio of a class is equal to the number of students of the class that will pass the exam divided by the total number of students of the class. The average pass ratio is the sum of pass ratios of all the classes divided by the number of the classes.
Return the maximum possible average pass ratio after assigning the extraStudents students. Answers within 10-5 of the actual answer will be accepted.
解題思路
初步分析
- 這題主要考察什麼概念?
- 貪心算法，找出每次分配學生，通過率可以提升至最高
- Pass Ratio = pass / total
- Average Pass Ratio = sum of pass ration / number of class
解法概述
-
暴力解法:
- 思路：
- 每次分配一個學生看哪個班級提升率會最高
- leetcode 這邊會 time limit exceeded
- 時間複雜度：O(N)
- 空間複雜度：O(N)
- 思路：
-
最佳解法:
- 思路：
- Priority Queue(Max Heap) 存每個班級潛在
- 時間複雜度：O(?)
- 空間複雜度：O(?)
- 思路：
測試案例
範例輸入輸出
Input: classes = [[1,2],[3,5],[2,2]], extraStudents = 2
Output: 0.78333
Explanation: You can assign the two extra students to the first class. The average pass ratio will be equal to (3/4 + 3/5 + 2/2) / 3 = 0.78333.
邊界情況
- 1 <= classes.length <= 10^5
- classes[i].length == 2
- 1 <= passi <= totali <= 10^5
- 1 <= extraStudents <= 10^5
學習筆記
今天學到什麼？
- 第一次使用 Priority Queue
- 第一次使用 Unit Test
遇到的困難
- 可能要再了解 Priority Queue 更多實際用途
改善方向
相關題目
- #1627 Graph Connectivity With Threshold
- #1728 Cat and Mouse II
- #3440 Reschedule Meetings for Maximum Free Time II
總結:
- 貪心策略，每次都選擇當下收益最大解
- 練習使用Priority Queue 快速找到最大收益